Questions

Solve the following Question.(1 Marks)

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
Evaluate the following Limits:

$\lim _{x \rightarrow 0}\left[\frac{\log (1+9 x)}{x}\right]$

Answer
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\log (1+9 x)}{x} \\
& =\lim _{x \rightarrow 0}\left[\frac{\log (1+9 x)}{9 x}\right] \times 9 \\
& =1 \times 9 \quad \ldots\left[\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right] \\
& =9 \quad \text { }
\end{aligned}
$
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Question 21 Mark
Evaluate the following Limits:

$\lim _{x \rightarrow 0}\left(1+\frac{x}{5}\right)^{\frac{1}{x}}$

Answer
$\begin{aligned} & \lim _{x \rightarrow 0}\left(1+\frac{x}{5}\right)^{\frac{1}{x}} \\ & =\lim _{x \rightarrow 0}\left[\left(1+\frac{x}{5}\right)^{\frac{5}{x}}\right]^{\frac{1}{5}} \text { } \\ & =\mathrm{e}^{\frac{1}{5}} \quad \ldots\left[\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=\mathrm{e}\right]\end{aligned}$
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Question 31 Mark
Evaluate the following Limits:

$\lim _{x \rightarrow 0}\left[\frac{5^x-1}{x}\right]$

Answer
$\lim _{x \rightarrow 0} \frac{5^x-1}{x}=\log 5 \quad \ldots\left[\lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
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Question 51 Mark
Evaluate the following Limits:$\lim _{x \rightarrow 2}\left[\frac{(x-2)}{2 x^2-7 x+6}\right]$
Answer
$ \lim _{x \rightarrow 2}\left(\frac{x-2}{2 x^2-7 x+6}\right)$
$ =\lim _{x \rightarrow 2} \frac{x-2}{(x-2)(2 x-3)}$
$ =\lim _{x \rightarrow 2} \frac{1}{2 x-3} \ldots [\text { As } x \rightarrow 2, x \neq 2 \therefore x-2 \neq 0]$
$ =\frac{1 \text { }}{2(2)-3}$
$ =1$
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Question 71 Mark
Evaluate the following Limits:

If $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$ then find the value of $\mathrm{n}$.

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Question 81 Mark
Evaluate the following limits : $\lim _{x \rightarrow 3}\left[\frac{x^2+2 x-15}{x^2-5 x+6}\right]$
Answer
$ \lim _{x \rightarrow 3}\left[\frac{x^2+2 x-15}{x^3-5 x+6}\right]$
$ =\lim _{x \rightarrow 3} \frac{(x+5)(x-3)}{(x-2)(x-3)}$
$ =\lim _{x \rightarrow 3} \frac{x+5}{x-2} \ldots\left[\begin{array}{l}\text { as } x \rightarrow 3, x \neq 3 \\ \therefore x-3 \neq 0\end{array}\right]$
$ =\frac{3+5}{3-2} \text { }$
$ =8$
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Question 91 Mark
Evaluate the following limits:

$\lim _{x \rightarrow 2}\left[\frac{x^3-4 x^2+4 x}{x^2-1}\right]$

Answer
$\begin{aligned} & \lim _{x \rightarrow 2}\left[\frac{x^3-4 x^2+4 x}{x^2-1}\right] \\ & =\lim _{x \rightarrow 2} \frac{x\left(x^2-4 x+4\right)}{\left(x^2-1\right)}=\lim _{x \rightarrow 2} \frac{x(x-2)^2}{x^2-1} \\ & =\frac{2(0)}{(2)^2-1}=\frac{2 \times 0}{3} \\ & =0 \text { }\end{aligned}$
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Question 151 Mark
Evaluate the following limits:

If $\lim _{x \rightarrow 1}\left[\frac{x^4-1}{x-1}\right]=\lim _{x \rightarrow a}\left[\frac{x^3-a^3}{x-a}\right]$, find all possible values of a.

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Question 161 Mark
Evaluate the following limits:

$\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]$

Answer
$\begin{aligned} \lim _{x \rightarrow 2} \frac{x^{-3}-2^{-3}}{x-2} & =(-3) \cdot(2)^{-4} \\ & \ldots\left[\because \lim _{x \rightarrow a} \frac{x^n-\mathrm{a}^{\mathrm{n}}}{x-\mathrm{a}}=\mathrm{na}^{\mathrm{n}-1}\right] \\ & =-3 \times \frac{1}{2^4} \\ & =\frac{-3}{16} \text { }\end{aligned}$
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Question 171 Mark
Evaluate the following limits:

$\lim _{x \rightarrow 3}\left[\frac{\sqrt{x+6}}{x}\right]$

Answer
$\begin{aligned} \lim _{x \rightarrow 3}\left[\frac{\sqrt{x+6}}{x}\right] & =\frac{\lim _{x \rightarrow 3} \sqrt{x+6}}{\lim _{x \rightarrow 3} x} \\ & =\frac{\sqrt{3+6}}{3} \\ & =\frac{\sqrt{9}}{3} \\ & =\frac{3}{3} \\ & =1\end{aligned}$
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